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Exercises 2.4 87
For the function defined by (9). both parts (i) and (iii) of the above definitionare true. See Figure 4. If 9 is the function defined by
1g(x) = - (x _ a)2
then both parts (ii) and (iv) are true, and the line x = a is a vertical asymptoteof the graph of g. This is shown in Figure 9.
y
EXA.MPLE 4 Find ihe vertical asymptote and draw a sketch of the graphof the function defined by
3f(x) =--3x-
Solution
3 'Jim --'= +:::1)
x-3+ X - 3I. 31m -- = -00
x-3-x-3
FIGURE 10It follows from Definition 2.4.8 that the line x = 3 is a verticai asymptote ofthe graph of f. A sketch of the graph of f appears in Figure 10.
EXERCISES 2.4
In Exercises I through J2, do the following: (a ) f.!.sea calculatorto tabulate values of f(x) for the specified values of x, and fromthese values make a statement regarding the apparent behallior off(x). (b) Find the indicated limit.
I . 0001. (a) f(x) = x _ 5; x IS 6, 5.5, 5.1, 5.01, 5.001, 5. I; "
. I(b) x~' x - 5
I2. (a) fix) = ;=5; x is 4, 4.5, 4.9, 4.99, 4.999, 4.9999;
I(b) }~~_ x _ 5
I .3. (a) f(x) = (x _ w; x IS 6, 5.5, 5.1, 5.01, 5.001,5.0001 and x
Iis 4, 4.5, 4.9, 4.99, 4.999, 4.9999; (b) lim ---2
.-5(x-5)x+2 .
4. (a) fix) = 1 _ x; x IS 0, 0.5, 0.9, 0.99. 0.999, 0.9999;
x+2(b) urn -1-
x ...•l " xx+2 .
5. (a) fix) = 1 _ x; x IS 2,1.5,1.1.1.01,1001,10001;
x+2(b) lim -I -
$-1" - X
x+26. (a) fix) = ---2; x is 0, 0.5, 0.9,0.99, 0.999, 0.9999 and x is
lx-I)
O 00 000 . x+22, 1.5, 1.1, 1. 1. 1. I, I. I; (b) lirn --2x-I (x - 1)
x - 2 .7. (a) fIx) = --; x IS 0, -0.5. -0.9, -0.99, -0.999,
x + Ix-2
-0.9999; (b) lim --x--I' x + I
x-28. (a) fix) = x + I; x is -2, -1.5, -1.1, - 1.01, -1.001,
x-2- 1.0001;(b) lim --
%-·,1- X +!
9. (a) fix) = _x_; x is -5, -4.5, -4.1, -4.01, -4.001,x+4
-4.0001; (b) lim __x_x- -4- X + 4
x .10. (a) fix) = --4;x IS 5. 4.5, 4.1, 4.01, 4.001, 4.0001;x-
(b) lim _x_x-4·x-4
4x11. (a) fIx) = ---2; x is -4, - 3.5, - 3.1, - 3.01, - 3.001,
9 - X'
4x-3.0001; (b) lim --2
x--3- 9 - x
-
I88 limits and Continuity
4X212. (a) fix) = --2; Xis 4, 3.5, 3.1, 3.01, 3.001, 3.0001;
9-x .4X2
(b) lim --2'x_"+9-x
1 J 1graph: (a) J\x) =;; (b) g(x) = ?; (c) hex) = 0;
. I(d) <p(x) = 4'
x34. For each of the following functions, find ill, verticalasyrnp,
tote of the graph of tile function, and draw a sketch of1 I I
the graph: (a) J(x) = --; ~b)g(x) = -2; (c) h(x) = -3;x x x
J(d)<p(x) = -4'
x
In Exercises 13 through 32, find the limit.
t + 2 -1+ 21}. lim -2-- 14. lim (-2)2
I-pt -4 1-2- r-
+ 2 '3 + X2. I' 1 16. lim _"_._is: rm -2--1-2-t -4 x-o- x
17. lim ,)3 + X2 18. lim.j) + X2
.x-o- X x-v .,!2
,)x2-9 ,~19. lim 20. lim v - x-x-p X - 3 x-4- X - 4. (1 1) xl - 32!. lim - - 2 22 lim -3--2x-O' ~ X r-O-X +x
23 lim 2 - 4x3
24. lim (_1 3_). x-o- 5r + 3r <-2- ,s - 2 S2 -'4
2S Jim ( 2 __ 3)• 1_ -4- 12 + 3r - 4 t + 4
2x3 _ 5x226. lim -.----:--
x-l- X2 - 1
In Exercises 35 through 42, find the verrical asympore(s) oj thegraph oj the Junction, and draw a sketch of the graph.
336.f(x) =-1x+
-438. fix) = --5x-
440. fix) = (x _ 5)2
.,35. f(x)=~4
x--2
37. fix) = --3x+-2
39. J(x) = (-3 2x + )
. 541. fix) = -X""'2-+-"8-x-+-1-:-5
1
42. fix) = 2 6x + 5x-
43. Prove that lim __ 3_2 = + CXJ by using Definition 2.4.1.x-2(x-2)
44. Prove that lim ~ = - CXJ by using Definition 2.4.2.x-4 (x - 4)
45. Prove Theorem 2.4.3(ii).47. Prove Theorem 2.4.4(iii).49. Prove Theorem 2.4.5.51. Prove Theorem 2.4.7.
, 15 - xl52. Use Definition 2.4.1 to prove that lim -- = -i- 00.<--33+y
[X2] - 128. lim ---x-,- X2 - I
6x2+-x-230. lim .,....-,2:-----
x--2·2x +3x-2x-2
32. lirn --' -===x-~- 2 - ,)4x - X2
21. lim [xl - xx-3- 3-x
x3 + 9x2 + 20x29. lim -..,,----:_=__x-3 x2+x-12
x-I31. lim ---=-
x-,·..)2x-x2-1
46. Prove Theorem 2.4.4(ii).48. Prove Theorem 2.4.4(iv).50. Prove Theorem 2.4.6.
33. For each of the following fun tions. find the vertical asymp-tote of the graph of the function. and draw a sketch of the
2.5 LIMiTS AT INFINITY The previous section was devoted ,0 infinite limits where function values eitherincreased or decreased without bound as the independent variable approacheda real number. Wc now consider limits of functions when the independentvariable either increases or decreases without bound. We begin with the func-tion defined by
Table 1---!-·-----·....-~.2
x fIx) = -;~--x- + 1
IY:J(x)=-~-
y2 + I(j
11.61.81.8823531.9230771.9XOI'lX1.9l)'!~O()1.'!999l}X
o1234 I
510
100I~ __ -
Let x take on the values O. I. 2.. -. -+.5. 10. 100. 1000, and so on, allowing x toincrease without hound. The eorr.:,;pontiing function values. either exact or ap-proximated by a calculator to six decimal places, are given in Table 1. Observefrom the table that as .v inCl'ca,cs through positive values, the function valuesget closer and closer to 2.
ominator by
e have
ntal asymptote.
d draw a sketch
e two functionsvalues of x for'on 1.5 and ex-
~u (2, + (0).
10
11I
!~. ----·2 ----1--------? Ii" I"10 2
_____ -2~--1-------~v11
bwcisa2.5 97
by Definition 2.4.8(1)the line x = 2 is a vertical asymptote of the graph of fl'
lim [,(x) ~ lim 2J I 2%-+00 %-+00 1 __
x
=2Thus by Definition 2.5.4(i)the line y = 2 is a horizontal asymptote of the graphofft.
Similarly, lim flex) = 2 A sketch of the graph of fl is in Figure 9.:
EJocises 1 through 10, do the following: Use a calculator tozie the values of f(x) for the specified values of x. (a) What:-'x) 'appear /0 be approaching as x increases without bound?
doesf(x) appear to be approaching as x decreases with·. (c) Find Jim f(x). (d) Find lim f(x),
x- +«1
ISES 2.5
= -00
Hence, by Definition 2.4.8(ii)the line x = 2 is a vertical asymptote of the graphof-f2' .
lim f2(x} = lim [-2) I 2]%-+co .x-+oo 1 __
x
= -2
Thus, by Definition 2.5.4(i}the line y = - 2 is a horizontai asymptote of thegraph of f2'
Also, lim f2(x) = - 2. A sketch of the graph of f2 appears in Figure 10.
The graph of the given equation is the union of the graphs of fl and f2' anda sketch is in Figure 11.
x) = ~2; x is 1,2,4,6,8,10,100, toOO and x is -1, -2,- x_j. -6, -8, -10, -100. -1(01).
_ xl =~; x is l, 2, 4, 6, 8, 10,.100,1000and x is -1. -2,x
_.!. -6, -8, -10, -100, -1000.
'-xl = ~; x is 1,2,4,6,8, 10, 100, 1000and x is -I, -2,x
_.!. -6. -8, -lO, -100, -1000.?
)= -.;; x is 1,2,4,6,8, to, 100,IC'.lOand x is -1, -2,x
_!. - 6. - 8, -10, -100, -1000.
S. f(x) = - ;X2 ; x is 0, 1, 2, 4, 6, 8, !O,100, 1000and x isx + 1 .
-1, -2, -4, -6, -8, -to, -100, -1000.x3
6. f(x) =-3--; x is 2, 4, 6, 8,10,100,1000 and x is -2, -4,x +2
-6, -8, -10, -100, -1000.4x + 17. f(x) = --; x is 2, 6, to, 100, 1000,10,000,100,000and2x - 1 .
x is - 2, - 6, - 10, - 100, - 1000,- to,OOO, - 100,000.5x - 3
8. f(x) = --; x is 2, 6,10, 100, 1000,to,OOC, 100,000andlOx + 1 .
x is -2, -6, -to, -100, -1000, -10,000, -100,000.
9. f(x) = x'~ 1; x is 2,6, 11),100,1000,10,000,100,000and x isx
-2, -6, -Hi, -100, -1000, -10,000, -100,000.
10. f(x) = ~; x is 2, 6, 10,100,1000,10,000,100,000and x isx+l
-2, -6, ..cl0, -100, -1000, -10,000, -100,000.
In Exercises 11 through 30, find the limit.
21 -e- jII. Jim .--
, __ o 5t - 2
2x + 713. Jim -;:--
x-« -" 4 - ox7x2 - 2x + 1
15. lim ---02,.-----x-+003x +8x+5
x+417. lim -2--
x_Tx.3x - S
lv' - 3y19. lim -'---
y_+'" y+ 14x3 + 2x' - 5
21. Jim -3-----x--'" 8x + x + 2
2v3 - 423. Jim -'---
y_+",Sy+3
25. x~~'" (3X + :,). ,,/x' +4
27. 11m ---,r-04 T: X + 4
,j'''''''''---=2-w-,-' -=-329. lim ------ w+S
limits and Continuity
6x - 412. lirn
X-4 '":7 3x + 1
i + 5x14. lim ---:;--
x_+.:c,2-.Jx
· 4S2 + 316. hm -.--
,--oc 2s' - 1
x' + 518. lim --J-
.1'-. 'S' X
· x' - 2x + 520. lim -;);----
,_.OX) 7.>: + x + 1
· 3x4- 7x' + 2
22. lirn -----.x-+ x 2x· + 1
5x3- 12x + 7
24. lim 4x' _ 1
26. lim (~- 4t), ....•-+.x. (
JX2 +428. lim ---
x--x x+4
I. Jy. + 130. trn ---
v __ co 2/ - 3
x'44. C/(x) = -'-. 4 - x'
2x45. hl.') = ------
6x' + llx - 10
. -\46. fIx) = r=;;====Jx' + 5x + 6
x48. h(x)=~
...':x' - 9
In Exercises 49 through 56. find the horizontal and vertical as-ymptotes and draw a sketch of the graph of the equation.
SO. 2xy + 4x - 3y + 6 = 052. xy' + 3y' - 9x = 054. 2xy' + 4y' - 3x = 0
49. 3xy - 2x - 4y - 3 = 051. x'y' - x' + 4y' = 053. (y' - l)(x·- 3) = 655. x' y .. 2x' - y - 2 = 056. x'y + 4xy - x' + x + 4y - 6 = 0
In Exercises 57 through 60, prove that Jim f(x) = 1 by applyingZ-+a;)
Definition 2.5.1; that is, for any € > 0 show that there exists anumber N > 0 such that ifx > N. then If(x) - 11 < e.
x 2x57. fIx) = x _ \ 58. f(x) = 2x + 3
x' - 1 X2 + 2x59. f(x) = X2 + 1 60. f(x) = ~
31. 111.1 (,. ::: + 1 - x]
In Exercises 31 through 36, find the limit (Pint: First obtain a[racti.n. \\ ii;; .1 rational numerat or.}
34. hrn I\/~~ I - x)x-" r: r- ----;:;::=..:....=
36. lirn ,./ t +..:! t + ..,fr
,-.'" .../1+1In Exercises 37 through 48. find the horizontal and vertical as-ymptotes and draw a sketch of the graph of the function.
,r ~ t ..,
2x + I37. fIx) = --1
x'· ,\
39. g(x) = 1 . -x
241. f(x) = ,-
,.\,··4
.'(- .•.. ~
4 - 3x38. {(x) =--
. .x + I
. \40. "(x) = : + "2
x
41 Ftx' - 3x. (.\')=-'=,'x' + 3
. 8x + 3 .61. Prove that 11m -- = 4 by showing that for any" > 0
x--0:2-,,-\there exists a number N < 0 such that if x < N then
1
8.>: +...2 - 41 < E.
2x - I62. Prove part (i) of Limit Theorem \2 (2:4.4) if "--; ....•a" is re-
placed by "x ....•+ "f."
63. Give a definition for each of the fellowing:(a) lim f(x) = - x; (b) lim f(x) = + 00;
x ....•+0:
(c) Jim f(x) = -"X:.70-- <T.
64. Prove that lim (x' - 4) = + 00 by showing that for any
N > 0 there exists an M > 0 such that if x > M thenx' - 4> N.
65. Prove that lim 16 - x - x') = - 00 by applying the de-x-+a:
finition in Exercise 63(a).66. Prove part (ii) of Limit Theorem 13 (2.5.3).
2.6 CONTINUITY OF AFUIVCTIOI\t AT A NUMilER
In Illustration I of Section 2.3 we discussed the function C de/hied by
C(X)={x ifO:s;x:s;lO (1)0.9x iflO<x
where C(x) dollars is the total cost of x pounds of a product. We showed thatJim C(x) does not exist because Jim C(x)"# Jim C(x). A sketch of the graph70-10 x-lO- x-to.
